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For this Question: You are given Q queries. In each query, determine whether there exists an array A of size N such that:
1. All the elements are positive integers.
2. The number of subarrays such that their XOR-sum is 0 is exactly K. In other words, there are exactly K pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ N and A<sub>l</sub> ⊕ A<sub>l+1</sub> ⊕ ... A<sub>r</sub> = 0.
If there exists such an array, print "YES", otherwise print "NO".The first line of the input contains a single integer Q — the number of queries (1 ≤ Q ≤ 10<sup>5</sup>).
Q lines follow, each line containing two space separated integers N (1 ≤ N ≤ 1000) and K (0 ≤ K ≤ N(N+1)/2).For each test case, print "YES", if there exists such an array, otherwise print "NO" (without the quotes).
Note that the output is case sensitive.Sample Input
3
2 2
3 2
2 1
Sample Output
NO
YES
YES, I have written this Solution Code: #include <bits/stdc++.h>
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e9;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
struct info {int n, k, idx;};
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
read(t);
vector<vector <info>> q(1001);
FOR (i, 1, t)
{
readb(n, k);
q[n/2 + 1].pb({n, k, i});
}
vi dp(1001*500 + 5, inf);
dp[0] = 0;
bool ans[t + 1] = {};
FOR (i, 1, 502)
{
FOR (j, i*(i - 1)/2, 1001*500)
dp[j] = min(dp[j], dp[j - i*(i - 1)/2] + i);
for (auto x: q[i])
if (dp[x.k] <= x.n + 1) ans[x.idx] = 1;
}
FOR (i, 1, t)
if (ans[i]) cout << "YES" << endl;
else cout << "NO" << endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Palindrome is a word, phrase, or sequence that reads the same backwards as forwards. Use recursion to check if a given string is palindrome or not.User Task:
Since this is a functional problem, you don't have to worry about the input, you just have to complete the function check_Palindrome() where you will get input string, starting index of string (which is 0) and the end index of string( which is str.length-1) as argument.
Constraints:
1 ≤ T ≤ 100
1 ≤ N ≤ 10000Return true if given string is palindrome else return falseSample Input
2
ab
aba
Sample Output
false
true, I have written this Solution Code:
static boolean check_Palindrome(String str,int s, int e)
{
// If there is only one character
if (s == e)
return true;
// If first and last
// characters do not match
if ((str.charAt(s)) != (str.charAt(e)))
return false;
// If there are more than
// two characters, check if
// middle substring is also
// palindrome or not.
if (s < e + 1)
return check_Palindrome(str, s + 1, e - 1);
return true;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Edward plays "Game 23". Initially, he has a number n and his goal is to transform it into m. In one move, he can multiply n by 2 or multiply n by 3. He can perform any number of moves. Print the number of moves needed to transform n to m. Print -1 if it is impossible to do so.The input consists of two space- separated integers n and m.
<b>Constraints</b>
1 ≤ n ≤ m ≤ 5⋅10<sup>8</sup>Print the number of moves to transform n to m, or -1 if there is no solution.<b>Sample Input 1</b>
120 51840
<b>Sample Output 1</b>
7
<b>Sample Input 2</b>
48 72
<b>Sample Output 2</b>
-1, I have written this Solution Code: #include <bits/stdc++.h>
#define ll long long
#define pb push_back
using namespace std;
int main()
{
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
ll n, m;
cin >> n >> m;
ll ans = 0;
if(m % n != 0){
cout << -1;
return 0;
}
else{
m /= n;
while(m % 3 == 0){
m /= 3;
ans++;
}
while(m % 2 == 0){
m /= 2;
ans++;
}
if(m == 1){
cout << ans;
}
else{
cout << -1;
}
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of even length N consisting of non-negative integers. You need to make each element of the array equal to zero, by doing some number of moves (possibly zero).
In a single move, you can choose two integers l and r such that 1 ≤ l ≤ r ≤ N. Let x = A[l] ⊕ A[l+1] ⊕ A[l+2] ... A[r], where ⊕ represents xor operator. Then, replace the whole subarray with x. In other words, assign A[i] = x for l ≤ i ≤ r.
Print the minimum number of moves required to make the entire array zero.The first line of the input contains a single integer T – the number of test cases.
Then T test cases follow, each test case in the following format:
The first line contains a single integer N – the length of array A.
The next line contains N space-separated integers representing the elements of the array A.
<b>Constraints:</b>
1≤ T ≤ 50
1 ≤ N ≤ 10<sup>3</sup>
0 ≤ A[i] ≤ 10<sup>9</sup>
N is even.Output T lines, the i<sup>th</sup> line containing a single integer – the answer to the i<sup>th</sup> test case.Sample Input:
2
2
2 2
4
7 1 2 0
Sample Output:
1
2, I have written this Solution Code: import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;
public class Main {
public static void main(String[] args) throws Exception {
Thread thread = new Thread(null, new TaskAdapter(), "@debanjandhar12", 1 << 29);
thread.start();
thread.join();
}
static class TaskAdapter implements Runnable {
@Override
public void run() {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
ArrayXor solver = new ArrayXor();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++)
solver.solve(i, in, out);
out.close();
}
}
static class ArrayXor {
public void solve(int testNumber, InputReader in, OutputWriter out) {
int n = in.readInt();
int[] a = in.readIntArray(n);
boolean ansIs0 = true;
for (int i = 0; i < n; i++) {
if (a[i] != 0) {
ansIs0 = false;
break;
}
}
if (ansIs0) {
out.printLine(0);
return;
}
int xor = 0;
for (int i = 0; i < n; i++) {
xor ^= a[i];
}
if (xor == 0) {
out.printLine(1);
return;
}
out.printLine(2);
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int[] readIntArray(int size) {
int[] array = new int[size];
for (int i = 0; i < size; i++) {
array[i] = readInt();
}
return array;
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int readInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return readString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void close() {
writer.close();
}
public void printLine(int i) {
writer.println(i);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of even length N consisting of non-negative integers. You need to make each element of the array equal to zero, by doing some number of moves (possibly zero).
In a single move, you can choose two integers l and r such that 1 ≤ l ≤ r ≤ N. Let x = A[l] ⊕ A[l+1] ⊕ A[l+2] ... A[r], where ⊕ represents xor operator. Then, replace the whole subarray with x. In other words, assign A[i] = x for l ≤ i ≤ r.
Print the minimum number of moves required to make the entire array zero.The first line of the input contains a single integer T – the number of test cases.
Then T test cases follow, each test case in the following format:
The first line contains a single integer N – the length of array A.
The next line contains N space-separated integers representing the elements of the array A.
<b>Constraints:</b>
1≤ T ≤ 50
1 ≤ N ≤ 10<sup>3</sup>
0 ≤ A[i] ≤ 10<sup>9</sup>
N is even.Output T lines, the i<sup>th</sup> line containing a single integer – the answer to the i<sup>th</sup> test case.Sample Input:
2
2
2 2
4
7 1 2 0
Sample Output:
1
2, I have written this Solution Code: for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
if a.count(0) == n:
print(0)
else:
c= 0
xor= 0
for x in a:
xor ^= x
if xor == 0:
print(1)
else:
print(2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of even length N consisting of non-negative integers. You need to make each element of the array equal to zero, by doing some number of moves (possibly zero).
In a single move, you can choose two integers l and r such that 1 ≤ l ≤ r ≤ N. Let x = A[l] ⊕ A[l+1] ⊕ A[l+2] ... A[r], where ⊕ represents xor operator. Then, replace the whole subarray with x. In other words, assign A[i] = x for l ≤ i ≤ r.
Print the minimum number of moves required to make the entire array zero.The first line of the input contains a single integer T – the number of test cases.
Then T test cases follow, each test case in the following format:
The first line contains a single integer N – the length of array A.
The next line contains N space-separated integers representing the elements of the array A.
<b>Constraints:</b>
1≤ T ≤ 50
1 ≤ N ≤ 10<sup>3</sup>
0 ≤ A[i] ≤ 10<sup>9</sup>
N is even.Output T lines, the i<sup>th</sup> line containing a single integer – the answer to the i<sup>th</sup> test case.Sample Input:
2
2
2 2
4
7 1 2 0
Sample Output:
1
2, I have written this Solution Code:
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define pi 3.141592653589793238
#define int long long
#define ll long long
#define ld long double
using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
long long powm(long long a, long long b,long long mod) {
long long res = 1;
while (b > 0) {
if (b & 1)
res = res * a %mod;
a = a * a %mod;
b >>= 1;
}
return res;
}
ll gcd(ll a, ll b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(0);
#ifndef ONLINE_JUDGE
if(fopen("input.txt","r"))
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
}
#endif
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
int a[n];
bool flag=true;
int ans=0;
for(int i=0;i<n;i++)
{
cin>>a[i];
ans^=a[i];
if(a[i])
flag=false;
}
if(flag)
cout<<0<<'\n';
else if(ans==0)
cout<<1<<'\n';
else
cout<<2<<'\n';
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: static void printDataTypes(int a, long b, float c, double d, char e)
{
System.out.println(a);
System.out.println(b);
System.out.printf("%.2f",c);
System.out.println();
System.out.printf("%.4f",d);
System.out.println();
System.out.println(e);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: void printDataTypes(int a, long long b, float c, double d, char e){
cout<<a<<endl;
cout<<b<<endl;
cout <<fixed<< std::setprecision(2) << c << '\n';
cout <<fixed<< std::setprecision(4) << d << '\n';
cout<<e<<endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: a=int(input())
b=int(input())
x=float(input())
g = "{:.2f}".format(x)
d=float(input())
e = "{:.4f}".format(d)
u=input()
print(a)
print(b)
print(g)
print(e)
print(u), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a large integer N. Find the sum of its digits. Eg:- if the integer is 1234, the answer is 1+2+3+4=10.The first and only line of input contains the integer N.
Constraints
The number of digits in N won't exceed 100000.Output a single integer, the sum of digits in N.Sample Input
1234
Sample Output
10
Sample Input
11111111111111111111
Sample Output
20, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str=br.readLine();
int sum=0;
for(int i=0;i<str.length();i++){
char c=str.charAt(i);
int k=c-'0';
sum+=k;}
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a large integer N. Find the sum of its digits. Eg:- if the integer is 1234, the answer is 1+2+3+4=10.The first and only line of input contains the integer N.
Constraints
The number of digits in N won't exceed 100000.Output a single integer, the sum of digits in N.Sample Input
1234
Sample Output
10
Sample Input
11111111111111111111
Sample Output
20, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
string s; cin>>s;
int ans = 0;
For(i, 0, sz(s)){
ans += (s[i]-'0');
}
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a large integer N. Find the sum of its digits. Eg:- if the integer is 1234, the answer is 1+2+3+4=10.The first and only line of input contains the integer N.
Constraints
The number of digits in N won't exceed 100000.Output a single integer, the sum of digits in N.Sample Input
1234
Sample Output
10
Sample Input
11111111111111111111
Sample Output
20, I have written this Solution Code: s = input()
count = 0
for x in s:count+=int(x)
print(count) ;, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: static void printString(String stringVariable){
System.out.println(stringVariable);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: S=input()
print(S), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: void printString(string s){
cout<<s;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: t=int(input())
while t!=0:
m,n=input().split()
m,n=int(m),int(n)
for i in range(m):
arr=input().strip()
if '1' in arr:
arr='1 '*n
else:
arr='0 '*n
print(arr)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define N 1000
int a[N][N];
// Driver code
int main()
{
int t;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
bool b[n];
for(int i=0;i<n;i++){
b[i]=false;
}
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
cin>>a[i][j];
if(a[i][j]==1){
b[i]=true;
}
}
}
for(int i=0;i<n;i++){
if(b[i]){
for(int j=0;j<m;j++){
cout<<1<<" ";
}}
else{
for(int j=0;j<m;j++){
cout<<0<<" ";
}
}
cout<<endl;
}
}}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws Exception{
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader bf = new BufferedReader(isr);
int t = Integer.parseInt(bf.readLine());
while (t-- > 0){
String inputs[] = bf.readLine().split(" ");
int m = Integer.parseInt(inputs[0]);
int n = Integer.parseInt(inputs[1]);
String[] matrix = new String[m];
for(int i=0; i<m; i++){
matrix[i] = bf.readLine();
}
StringBuffer ones = new StringBuffer("");
StringBuffer zeros = new StringBuffer("");
for(int i=0; i<n; i++){
ones.append("1 ");
zeros.append("0 ");
}
for(int i=0; i<m; i++){
if(matrix[i].contains("1")){
System.out.println(ones);
}else{
System.out.println(zeros);
}
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
try{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int testcase = Integer.parseInt(br.readLine());
for(int t=0;t<testcase;t++){
int num = Integer.parseInt(br.readLine().trim());
if(num==1)
System.out.println("No");
else if(num<=3)
System.out.println("Yes");
else{
if((num%2==0)||(num%3==0))
System.out.println("No");
else{
int flag=0;
for(int i=5;i*i<=num;i+=6){
if(((num%i)==0)||(num%(i+2)==0)){
System.out.println("No");
flag=1;
break;
}
}
if(flag==0)
System.out.println("Yes");
}
}
}
}catch (Exception e) {
System.out.println("I caught: " + e);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: t=int(input())
for i in range(t):
number = int(input())
if number > 1:
i=2
while i*i<=number:
if (number % i) == 0:
print("No")
break
i+=1
else:
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
long long n,k;
cin>>n;
long x=sqrt(n);
int cnt=0;
vector<int> v;
for(long long i=2;i<=x;i++){
if(n%i==0){
cout<<"No"<<endl;
goto f;
}}
cout<<"Yes"<<endl;
f:;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Your task is to find the number of days in the given non-leap year month given the month's number.
<b>Example</b>
Take input as 3 which refers to the 3rd month which is March.
Hence the output must be 31 as there are 31 days in March month.User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>numberofdays()</b> which contains M as a parameter.
<b>Constraints:- </b>
1 ≤ M ≤ 12Print the number of days in the particular month.Sample Input 1:-
1
Sample Output 1:
31
Sample Input 2:-
2
Sample Output 2:-
28, I have written this Solution Code: static void numberofdays(int M){
if(M==4 || M ==6 || M==9 || M==11){System.out.print(30);}
else if(M==2){System.out.print(28);}
else{
System.out.print(31);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to become rich so he came up with an idea, So, he buys some gadgets from the future at a price of C and sells them at a price of S to his friends. Now Nobita wants to know how much he gains by selling all gadget. As we all know Nobita is weak in maths help him to find the profit he getsYou don't have to worry about the input, you just have to complete the function <b>Profit()</b>
<b>Constraints:-</b>
1 <= C <= S <= 1000Print the profit Nobita gets from selling one gadget.Sample Input:-
3 5
Sample Output:-
2
Sample Input:-
9 16
Sample Output:-
7, I have written this Solution Code: def profit(C, S):
print(S - C), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to become rich so he came up with an idea, So, he buys some gadgets from the future at a price of C and sells them at a price of S to his friends. Now Nobita wants to know how much he gains by selling all gadget. As we all know Nobita is weak in maths help him to find the profit he getsYou don't have to worry about the input, you just have to complete the function <b>Profit()</b>
<b>Constraints:-</b>
1 <= C <= S <= 1000Print the profit Nobita gets from selling one gadget.Sample Input:-
3 5
Sample Output:-
2
Sample Input:-
9 16
Sample Output:-
7, I have written this Solution Code: static void Profit(int C, int S){
System.out.println(S-C);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main (String[] args)throws IOException {
Reader sc = new Reader();
int N = sc.nextInt();
int[] arr = new int[N];
for(int i=0;i<N;i++){
arr[i] = sc.nextInt();
}
int max=0;
if(arr[0]<arr[N-1])
System.out.print(N-1);
else{
for(int i=0;i<N-1;i++){
int j = N-1;
while(j>i){
if(arr[i]<arr[j]){
if(max<j-i){
max = j-i;
} break;
}
j--;
}
if(i==j)
break;
if(j==N-1)
break;
}
if(max==0)
System.out.print("-1");
else
System.out.print(max);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
/* For a given array arr[],
returns the maximum j – i such that
arr[j] > arr[i] */
int maxIndexDiff(int arr[], int n)
{
int maxDiff;
int i, j;
int *LMin = new int[(sizeof(int) * n)];
int *RMax = new int[(sizeof(int) * n)];
/* Construct LMin[] such that
LMin[i] stores the minimum value
from (arr[0], arr[1], ... arr[i]) */
LMin[0] = arr[0];
for (i = 1; i < n; ++i)
LMin[i] = min(arr[i], LMin[i - 1]);
/* Construct RMax[] such that
RMax[j] stores the maximum value from
(arr[j], arr[j+1], ..arr[n-1]) */
RMax[n - 1] = arr[n - 1];
for (j = n - 2; j >= 0; --j)
RMax[j] = max(arr[j], RMax[j + 1]);
/* Traverse both arrays from left to right
to find optimum j - i. This process is similar to
merge() of MergeSort */
i = 0, j = 0, maxDiff = -1;
while (j < n && i < n)
{
if (LMin[i] < RMax[j])
{
maxDiff = max(maxDiff, j - i);
j = j + 1;
}
else
i = i + 1;
}
return maxDiff;
}
// Driver Code
signed main()
{
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
int maxDiff = maxIndexDiff(a, n);
cout << maxDiff;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: n=int(input())
arr=list(map(int,input().split()))
rightMax = [0] * n
rightMax[n - 1] = arr[n - 1]
for i in range(n - 2, -1, -1):
rightMax[i] = max(rightMax[i + 1], arr[i])
maxDist = -2**31
i = 0
j = 0
while (i < n and j < n):
if (rightMax[j] >= arr[i]):
maxDist = max(maxDist, j - i)
j += 1
else:
i += 1
if maxDist==0:
maxDist=-1
print(maxDist), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: static void printString(String stringVariable){
System.out.println(stringVariable);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: S=input()
print(S), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: void printString(string s){
cout<<s;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement Selection Sort on a given array, and make it sorted.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in ascending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=10000Sorted output where each element is space separatedSample Input:
3
3 1 2
Sample Output:
1 2 3, I have written this Solution Code: // arr is unsorted array
// n is the number of elements in the array
function selectionSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => a - b)
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement Selection Sort on a given array, and make it sorted.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in ascending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=10000Sorted output where each element is space separatedSample Input:
3
3 1 2
Sample Output:
1 2 3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
String[] str = br.readLine().split(" ");
int a[] = new int[n];
for(int i=0; i<n; i++){
a[i] = Integer.parseInt(str[i]);
}
for(int i=0; i<n-1; i++){
for(int j=i+1; j<n; j++){
if(a[i]>a[j]){
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
for(int i=0; i<n; i++){
System.out.print(a[i]+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement Selection Sort on a given array, and make it sorted.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in ascending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=10000Sorted output where each element is space separatedSample Input:
3
3 1 2
Sample Output:
1 2 3, I have written this Solution Code: def selectionSort(arr):
arr.sort()
return arr
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement Selection Sort on a given array, and make it sorted.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in ascending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=10000Sorted output where each element is space separatedSample Input:
3
3 1 2
Sample Output:
1 2 3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n);
for(int i=0;i<n;i++){
cout<<a[i]<<" ";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N. The task is to find the square root of N. If N is not a perfect square, then return floor(√N). Try to solve the problem using Binary Search.The first line of input contains the number of test cases T. For each test case, the only line contains the number N.
<b>Constraints:</b>
1 ≤ T ≤ 10000
0 ≤ x ≤ 10<sup>8</sup>For each testcase, print square root of given integer.Sample Input:
2
5
4
Sample Output:
2
2
<b>Explanation:</b.
Testcase 1: Since, 5 is not a perfect square, the floor of square_root of 5 is 2.
Testcase 2: Since 4 is a perfect square, its square root is 2., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int binarySearch( int k,int start,int end) {
int mid = start + (end - start)/2;
if (mid * mid == k) {
return (int) mid;
}
else if ((mid*mid) > end) {
return binarySearch( k, 0, mid - 1);
}
else if ((mid * mid < end)) {
return binarySearch( k, mid + 1, end);
}
else {
mid = (int) Math.sqrt(k);
}
return mid;
}
public static void main (String[] args) throws IOException{
InputStreamReader st = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(st);
int T = Integer.parseInt(br.readLine());
while (T-- > 0) {
int n = Integer.parseInt(br.readLine());
int k = n;
System.out.println(binarySearch( k, 0, n - 1));
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N. The task is to find the square root of N. If N is not a perfect square, then return floor(√N). Try to solve the problem using Binary Search.The first line of input contains the number of test cases T. For each test case, the only line contains the number N.
<b>Constraints:</b>
1 ≤ T ≤ 10000
0 ≤ x ≤ 10<sup>8</sup>For each testcase, print square root of given integer.Sample Input:
2
5
4
Sample Output:
2
2
<b>Explanation:</b.
Testcase 1: Since, 5 is not a perfect square, the floor of square_root of 5 is 2.
Testcase 2: Since 4 is a perfect square, its square root is 2., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define ll long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
int sqr(int a) {
int A=a;
if(A<2) return A;
ll l=1,r=A;
ll k=1;
while(l<=r)
{
ll mid=(l+r)/2;
ll u=mid*mid;
if(u<=A)
{
k=max(mid,k);
l=mid+1;
}else
{
r=mid-1;
}
}
return k;
}
signed main()
{
int t;
cin>>t;
while(t>0)
{
t--;
long a;
cin>>a;
long x = sqrt(a);
cout<<x<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N. The task is to find the square root of N. If N is not a perfect square, then return floor(√N). Try to solve the problem using Binary Search.The first line of input contains the number of test cases T. For each test case, the only line contains the number N.
<b>Constraints:</b>
1 ≤ T ≤ 10000
0 ≤ x ≤ 10<sup>8</sup>For each testcase, print square root of given integer.Sample Input:
2
5
4
Sample Output:
2
2
<b>Explanation:</b.
Testcase 1: Since, 5 is not a perfect square, the floor of square_root of 5 is 2.
Testcase 2: Since 4 is a perfect square, its square root is 2., I have written this Solution Code: N=int(input())
for i in range(0,N):
M=int(input())
s=M**0.5
print(int(s)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N. The task is to find the square root of N. If N is not a perfect square, then return floor(√N). Try to solve the problem using Binary Search.The first line of input contains the number of test cases T. For each test case, the only line contains the number N.
<b>Constraints:</b>
1 ≤ T ≤ 10000
0 ≤ x ≤ 10<sup>8</sup>For each testcase, print square root of given integer.Sample Input:
2
5
4
Sample Output:
2
2
<b>Explanation:</b.
Testcase 1: Since, 5 is not a perfect square, the floor of square_root of 5 is 2.
Testcase 2: Since 4 is a perfect square, its square root is 2., I have written this Solution Code: // n is the input number
function sqrt(n) {
// write code here
// do not console.log
// return the number
return Math.floor(Math.sqrt(n))
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, I have written this Solution Code: N = int(input())
Nums = list(map(int,input().split()))
f = False
for n in Nums:
if n < 0:
f = True
break
if (f):
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a;
bool win=false;
for(int i=0;i<n;i++){
cin>>a;
if(a<0){win=true;}}
if(win){
cout<<"Yes";
}
else{
cout<<"No";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i] = sc.nextInt();
}
for(int i=0;i<n;i++){
if(a[i]<0){System.out.print("Yes");return;}
}
System.out.print("No");
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Hi, it's Monica!
Monica looks at her FRIENDS circle and wonders if her circle is bigger than yours. Please let her know if her friends' circle is bigger than yours, given she has a friends' circle of size 6.The first and the only line of input contains a single integer N, the size of your friends' circle.
Constraints
1 <= N <= 10Output "Yes" if the size of Monica's friends circle has more friends than yours, else output "No".Sample Input
3
Sample Output
Yes
Sample Input
10
Sample Output
No, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Cf cf = new Cf();
cf.solve();
}
static class Cf {
InputReader in = new InputReader(System.in);
OutputWriter out = new OutputWriter(System.out);
int mod = (int)1e9+7;
public void solve() {
int t = in.readInt();
if(t>=6) {
out.printLine("No");
}else {
out.printLine("Yes");
}
}
public long findPower(long x,long n) {
long ans = 1;
long nn = n;
while(nn>0) {
if(nn%2==1) {
ans = (ans*x) % mod;
nn-=1;
}else {
x = (x*x)%mod;
nn/=2;
}
}
return ans%mod;
}
public static int log2(int x) {
return (int) (Math.log(x) / Math.log(2));
}
private static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int readInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public double readDouble() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, readInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, readInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public long readLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return readString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
private static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
writer.flush();
}
public void printLine(Object... objects) {
print(objects);
writer.println();
writer.flush();
}
public void close() {
writer.close();
}
public void flush() {
writer.flush();
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Hi, it's Monica!
Monica looks at her FRIENDS circle and wonders if her circle is bigger than yours. Please let her know if her friends' circle is bigger than yours, given she has a friends' circle of size 6.The first and the only line of input contains a single integer N, the size of your friends' circle.
Constraints
1 <= N <= 10Output "Yes" if the size of Monica's friends circle has more friends than yours, else output "No".Sample Input
3
Sample Output
Yes
Sample Input
10
Sample Output
No, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
template<class C> void mini(C&a4, C b4){a4=min(a4,b4);}
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define mod 1000000007ll
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
if(n<6)
cout<<"Yes";
else
cout<<"No";
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Hi, it's Monica!
Monica looks at her FRIENDS circle and wonders if her circle is bigger than yours. Please let her know if her friends' circle is bigger than yours, given she has a friends' circle of size 6.The first and the only line of input contains a single integer N, the size of your friends' circle.
Constraints
1 <= N <= 10Output "Yes" if the size of Monica's friends circle has more friends than yours, else output "No".Sample Input
3
Sample Output
Yes
Sample Input
10
Sample Output
No, I have written this Solution Code: n=int(input())
if(n>=6):
print("No")
else:
print("Yes"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Internet download speed is often expressed in bit per second whereas file size is expressed in Bytes. It is known that 1 Byte = 8 bits. Given file size in megabytes(MB) and internet speed in megabits per seconds(Mbps), find the time taken in seconds to download the file.The only line of input contains two integers denoting the file size in MB and download speed in Mbps.
1 <= file size <= 1000
1 <= download speed <= 1000Print a single integer denoting the time taken to download the file in seconds.
It is guaranteed that the result will be an integer.Sample Input:
10 16
Sample Output:
5, I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
void solve(){
int f, s;
cin >> f >> s;
cout << (8*f)/s << endl;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Internet download speed is often expressed in bit per second whereas file size is expressed in Bytes. It is known that 1 Byte = 8 bits. Given file size in megabytes(MB) and internet speed in megabits per seconds(Mbps), find the time taken in seconds to download the file.The only line of input contains two integers denoting the file size in MB and download speed in Mbps.
1 <= file size <= 1000
1 <= download speed <= 1000Print a single integer denoting the time taken to download the file in seconds.
It is guaranteed that the result will be an integer.Sample Input:
10 16
Sample Output:
5, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int f = sc.nextInt();
int s = sc.nextInt();
int ans = (8*f)/s;
System.out.print(ans);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Internet download speed is often expressed in bit per second whereas file size is expressed in Bytes. It is known that 1 Byte = 8 bits. Given file size in megabytes(MB) and internet speed in megabits per seconds(Mbps), find the time taken in seconds to download the file.The only line of input contains two integers denoting the file size in MB and download speed in Mbps.
1 <= file size <= 1000
1 <= download speed <= 1000Print a single integer denoting the time taken to download the file in seconds.
It is guaranteed that the result will be an integer.Sample Input:
10 16
Sample Output:
5, I have written this Solution Code: a = list(map(int,input().strip().split()))[:2]
print(int((a[0]*8)/a[1])), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, and an integer K your task is to find the lexicographically largest string by removing no more than K characters.The first line of input contains a single string S, and the next line of input contains a single integer K.
<b>Constraints:-</b>
1 <= |S| <= 10<sup>5</sup>
1 <= K <= 10<sup>5</sup>
Note:- String will contain only lowercase English letters.Print the lexicographically largest string possible after removing not more than K characters.Sample Input 1:-
abzccd
4
Sample Output 1:-
zd
Sample Input 2:-
dcba
5
Sample Output 2:-
dcba, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
signed main(){
// freopen("ou.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
string s;
cin>>s;
int k;
cin>>k;
int n = s.length();
queue<int> q[26];
for(int i=0;i<n;i++){
q[s[i]-'a'].push(i);
}
int i=0;
string ans="";
while(i<n){
// out1(i);out(ans);
for(int j=26;j>=0;j--){
while(q[j].size()!=0 && q[j].front()<i){
q[j].pop();
}
if(q[j].size() !=0 && q[j].front()-i<=k){
ans+=(char)((int)'a'+j);
k-=q[j].front()-i;
i=q[j].front()+1;
q[j].pop();
break;
}
}
}
out(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, and an integer K your task is to find the lexicographically largest string by removing no more than K characters.The first line of input contains a single string S, and the next line of input contains a single integer K.
<b>Constraints:-</b>
1 <= |S| <= 10<sup>5</sup>
1 <= K <= 10<sup>5</sup>
Note:- String will contain only lowercase English letters.Print the lexicographically largest string possible after removing not more than K characters.Sample Input 1:-
abzccd
4
Sample Output 1:-
zd
Sample Input 2:-
dcba
5
Sample Output 2:-
dcba, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner inp = new Scanner(System.in);
String str = inp.nextLine();
int n = inp.nextInt();
Stack<Character> st = new Stack<>();
int check = 0;
for(int i = 0 ; i < str.length() ; i++){
if(n > 0){
while(!st.isEmpty() && n > 0 && str.charAt(i) > st.peek()){
st.pop();
n--;
}
st.push(str.charAt(i));
check++;
}
else
continue;
}
String ans = "";
while(!st.isEmpty()){
ans = st.pop() + ans;
}
if(check < str.length()){
ans += str.substring(check, str.length());
}
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A of N integers, find the pair of integers in the array which have minimum XOR value. Report the minimum XOR value.First line denotes N, the size of the array.
Next line denotes N space-separated array elements.
Constraints:
2 <= N <= 100000
0 <= A[i] <= 10^7Print a single integer denoting minimum xor valueSample Input
4
0 2 5 7
Sample Output
2
Explanation:
0 xor 2 = 2
Sample Input
4
0 4 7 9
Sample Output
3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
int ans = 0;
int mini = Integer.MAX_VALUE;
Scanner sc = new Scanner(System.in);
int array_size = sc.nextInt();
int N[] = new int[array_size];
for (int i = 0; i < array_size; i++) {
if(mini == 0){
break;
}
N[i] = sc.nextInt();
for (int j = i - 1; j >= 0; j--) {
ans = N[i] ^ N[j];
if (mini > ans) {
mini = ans;
}
}
}
System.out.println(mini);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A of N integers, find the pair of integers in the array which have minimum XOR value. Report the minimum XOR value.First line denotes N, the size of the array.
Next line denotes N space-separated array elements.
Constraints:
2 <= N <= 100000
0 <= A[i] <= 10^7Print a single integer denoting minimum xor valueSample Input
4
0 2 5 7
Sample Output
2
Explanation:
0 xor 2 = 2
Sample Input
4
0 4 7 9
Sample Output
3, I have written this Solution Code: a=int(input())
lis = list(map(int,input().split()))
val=666666789
for i in range(a+1):
for j in range(i+1,a):
temp = lis[i]^lis[j]
if(temp<val):
val = temp
if(val==0):
break
print(val), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A of N integers, find the pair of integers in the array which have minimum XOR value. Report the minimum XOR value.First line denotes N, the size of the array.
Next line denotes N space-separated array elements.
Constraints:
2 <= N <= 100000
0 <= A[i] <= 10^7Print a single integer denoting minimum xor valueSample Input
4
0 2 5 7
Sample Output
2
Explanation:
0 xor 2 = 2
Sample Input
4
0 4 7 9
Sample Output
3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
// Function to find minimum XOR pair
int minXOR(int arr[], int n)
{
// Sort given array
sort(arr, arr + n);
int minXor = INT_MAX;
int val = 0;
// calculate min xor of consecutive pairs
for (int i = 0; i < n - 1; i++) {
val = arr[i] ^ arr[i + 1];
minXor = min(minXor, val);
}
return minXor;
}
// Driver program
int main()
{
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)
{
cin>>arr[i];
}
cout << minXOR(arr, n) << endl;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print the pattern of "*" in the form of the Right Angle Triangle.
See the below example for clarity.Since this is a functional problem, you don't have to worry about the input. It will be handled by driver code. You just have to complete <b>printTriangle()</b>.
In the custom input area, you can provide any positive integer and check whether your code is working or not.Print the right angle triangle of height 5 as shown.Sample Input:
No Input
Sample Output:-
*
* *
* * *
* * * *
* * * * *, I have written this Solution Code: class Solution {
public static void printTriangle(){
System.out.println("*");
System.out.println("* *");
System.out.println("* * *");
System.out.println("* * * *");
System.out.println("* * * * *");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print the pattern of "*" in the form of the Right Angle Triangle.
See the below example for clarity.Since this is a functional problem, you don't have to worry about the input. It will be handled by driver code. You just have to complete <b>printTriangle()</b>.
In the custom input area, you can provide any positive integer and check whether your code is working or not.Print the right angle triangle of height 5 as shown.Sample Input:
No Input
Sample Output:-
*
* *
* * *
* * * *
* * * * *, I have written this Solution Code: j=1
for i in range(0,5):
for k in range(0,j):
print("*",end=" ")
if(j<=4):
print()
j=j+1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a tree consisting of N vertices.
We define a K- star as a pair of:
- a center vertex C
- an unordered K- tuple of corner vertices ( v<sub>1</sub>,..... , v<sub>K</sub>)
which satisfies the following conditions for all (i, j) such that 1 <= i, j <= K and i != j:
- d(C, v<sub>i</sub>) = d(C, v<sub>j</sub>) != 0
- 2 * d(C, v<sub>i</sub>) = d(v<sub>i</sub>, v<sub>j</sub>)
Here d(x, y) denotes the distance between nodes x and y on the given tree.
Given the value K, count the number of K- stars in the given tree. Since this number may be huge, print the answer modulo 998244353.The first line of the input contains two integers: N, the number of nodes in the tree; and K, the star parameter.
It is followed by N-1 lines each containing two integers, u<sub>i</sub> and v<sub>i</sub>, the endpoints of an edge of the tree. It is guaranteed that the given edges form a tree.
<b>Constraints: </b>
3 <= N <= 5000
2 <= K <= N-1
1 <= u<sub>i</sub>, v<sub>i</sub> <= NPrint a single integer, the number of K- stars in the given tree, modulo 998244353.Sample Input 1:
4 3
1 2
1 3
1 4
Sample Output 1:
1
Sample Input 2:
7 2
1 2
1 3
2 4
2 5
3 6
3 7
Sample Output 2:
11
Sample Explanations:
In the first sample, there is exactly 1 3- star in the tree, centered at node 1 and with all other nodes being the corner vertices of the star.
In the second sample, there are five 2- stars centered at node 1, and three each centered on nodes 2 and 3. So, answer is 5 + 3 + 3 = 11., I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// #include <sys/resource.h>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
// PRAGMAS (do these even work?)
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
// DEFINE STATEMENTS
const long long infty = 1e18;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define pob pop_back
#define f first
#define s second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
#define start_clock() auto start_time = std::chrono::high_resolution_clock::now();
#define measure() auto end_time = std::chrono::high_resolution_clock::now(); cerr << (end_time - start_time)/std::chrono::milliseconds(1) << "ms" << endl;
typedef long long ll;
typedef long double ld;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
// DEBUG FUNCTIONS
#ifdef LOCALY
template<typename T>
void __p(T a) {
cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
cout<<"{";
__p(a.first);
cout<<",";
__p(a.second);
cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
cout<<"{";
for(auto it=a.begin(); it<a.end(); it++)
__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T>
void __p(std::multiset<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
cout<<"{\n";
for(auto it=a.begin(); it!=a.end();++it)
{
__p(it->first);
cout << ": ";
__p(it->second);
cout<<"\n";
}
cout << "}\n";
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
__p(a1);
__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
cout<<name<<" : ";
__p(arg1);
cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
int bracket=0,i=0;
for(;; i++)
if(names[i]==','&&bracket==0)
break;
else if(names[i]=='(')
bracket++;
else if(names[i]==')')
bracket--;
const char *comma=names+i;
cout.write(names,comma-names)<<" : ";
__p(arg1);
cout<<" | ";
__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif
// DEBUG FUNCTIONS END
// CUSTOM HASH TO SPEED UP UNORDERED MAP AND TO AVOID FORCED CLASHES
struct custom_hash {
static uint64_t splitmix64(uint64_t x) {
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
return splitmix64(x + FIXED_RANDOM);
}
};
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count()); // FOR RANDOM NUMBER GENERATION
ll mod_exp(ll a, ll b, ll c)
{
ll res=1; a=a%c;
while(b>0)
{
if(b%2==1)
res=(res*a)%c;
b/=2;
a=(a*a)%c;
}
return res;
}
ll mymod(ll a,ll b)
{
return (((a = a%b) < 0) ? a + b : a);
}
ll gcdExtended(ll,ll,ll *,ll *);
ll modInverse(ll a, ll m)
{
ll x, y;
ll g = gcdExtended(a, m, &x, &y);
g++; //this line was added just to remove compiler warning
ll res = (x%m + m) % m;
return res;
}
ll gcdExtended(ll a, ll b, ll *x, ll *y)
{
if (a == 0)
{
*x = 0, *y = 1;
return b;
}
ll x1, y1;
ll gcd = gcdExtended(b%a, a, &x1, &y1);
*x = y1 - (b/a) * x1;
*y = x1;
return gcd;
}
//Dont forget to reset MOD if some other MOD is needed
const ll MOD = num2;
//Comment the line above and uncomment the line below if problem requires more than 1 MOD
//After uncommenting the below line, declaration of Mint becomes [ Mint<mod> M; ]
//template<ll MOD>
class Mint
{
//WARNING:
//Be very careful not to use two Mints with different mods for any operation
//No guarantee of behavior in this case
public:
ll val;
static ll mod_exp(ll a, ll b){ ll res=1; a=a%MOD; while(b>0){ if(b%2==1) res=(res*a)%MOD; b/=2; a=(a*a)%MOD; } return res; }
static ll gcdExtended(ll a, ll b, ll *x, ll *y) { if (a == 0) { *x = 0, *y = 1; return b; } ll x1, y1; ll gcd = gcdExtended(b%a, a, &x1, &y1);*x = y1 - (b/a) * x1; *y = x1; return gcd; }
static ll modInverse(ll a) { ll x, y; ll g = gcdExtended(a, MOD, &x, &y); g++; ll res = (x%MOD); if(res < 0) res += MOD; return res;}
Mint(){ val = 0;}
Mint(ll x){ val = x%MOD; if(val < 0) val += MOD;}
Mint& operator +=(const Mint &other){ val += other.val; if(val >= MOD) val -= MOD; return (*this); }
Mint& operator -=(const Mint &other){ val -= other.val;if(val < 0) val += MOD; return (*this); }
Mint& operator *=(const Mint &other){ val = (val * other.val)%MOD; return (*this); }
Mint& operator /=(const Mint &other){ val = (val * modInverse(other.val)) % MOD; return (*this); }
Mint& operator =(const Mint &other) { val = other.val; return (*this); }
Mint operator +(const Mint &other) const { return Mint(*this) += other; }
Mint operator -(const Mint &other) const { return Mint(*this) -= other; }
Mint operator *(const Mint &other) const { return Mint(*this) *= other; }
Mint operator /(const Mint &other) const { return Mint(*this) /= other; }
bool operator ==(const Mint &other) const { return val == other.val; }
Mint operator ++() { ++val; if(val == MOD) val = 0; return (*this); }
Mint operator ++(int) { val++; if(val == MOD) val = 0; return Mint(val-1); }
Mint operator --() { --val; if(val == -1) val = MOD-1; return (*this); }
Mint operator --(int) { val--; if(val == -1) val = MOD-1; return Mint(val+1); }
// ^ has very low precedence, careful!!
template<typename T>
Mint& operator ^=(const T &other){ val = mod_exp(val, other); return (*this); }
template<typename T>
Mint operator ^(const T &other) const { return Mint(*this) ^= other; }
Mint& operator ^=(const Mint &other){ val = mod_exp(val, other.val); return (*this); }
Mint operator ^(const Mint &other) const { return Mint(*this) ^= other; }
template<typename T>
explicit operator T() { return (T)val; }
template<typename T>
friend Mint operator +(T other, const Mint &M){ return Mint(other) + M; }
template<typename T>
friend Mint operator -(T other, const Mint &M){ return Mint(other) - M; }
template<typename T>
friend Mint operator *(T other, const Mint &M){ return Mint(other) * M; }
template<typename T>
friend Mint operator /(T other, const Mint &M){ return Mint(other) / M; }
template<typename T>
friend Mint operator ^(T other, const Mint &M){ return Mint(other) ^ M; }
friend std::ostream &operator << (std::ostream &output, const Mint &M){ return output << M.val; }
friend std::istream &operator >> (std::istream &input, Mint &M) { input >> M.val; M.val %= MOD; return input;}
};
const int MAXN = 5e3+5;
Mint facto[MAXN], invfacto[MAXN];
void pre()
{
facto[0] = 1;
for(int i=1; i<MAXN; i++) facto[i] = i*facto[i-1];
invfacto[MAXN-1] = 1/facto[MAXN-1];
for(int i=MAXN-2; i>=0; i--) invfacto[i] = (i+1)*invfacto[i+1];
}
Mint nCr(ll a, ll b)
{
if(b < 0 || b > a) return Mint(0);
return (facto[a]*invfacto[b]*invfacto[a-b]);
}
struct Graph
{
vector<vector<int>> adj;
Graph(int n)
{
adj.resize(n+1);
}
void add_edge(int a, int b, bool directed = false)
{
adj[a].pb(b);
if(!directed) adj[b].pb(a);
}
Mint solve(int root, int k)
{
Mint ways = 0;
vector<vector<Mint>> parts(adj.size());
vector<bool> vis(adj.size(), false);
vis[root] = true;
for(auto &st: adj[root])
{
queue<pair<int, int>> q;
q.push({st, 0});
vis[st] = true;
int last = -1;
while(!q.empty())
{
pair<int, int> f = q.front();
if(f.second != last) {
parts[f.second].pb(1);
}
else {
parts[f.second].back()++;
}
last = f.second;
q.pop();
for(auto &p: adj[f.first]) {
if(!vis[p]) {
q.push({p, f.second+1});
vis[p] = true;
}
}
}
}
for(int i=0; i<parts.size(); i++) {
if(parts[i].size() >= k) {
vector<vector<Mint>> dp(parts[i].size(), vector<Mint>(k+1));
dp[0][0] = 1;
dp[0][1] = parts[i][0];
for(int j=1; j<parts[i].size(); j++)
{
dp[j][0] = 1;
for(int x = 1; x <= k; x++) {
//dont take current element
dp[j][x] += dp[j-1][x];
//take it
dp[j][x] += dp[j-1][x-1] * parts[i][j];
}
}
// trace(dp);
ways += dp[parts[i].size()-1][k];
}
else {
break;
}
}
return ways;
}
};
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n, k;
cin >> n >> k;
Mint ans = 0;
Graph G(n);
REP(i, 1, n)
{
ll a, b;
cin >> a >> b;
G.add_edge(a, b);
}
REP(i, 1, n+1)
{
if(G.adj[i].size() >= k) {
ans += G.solve(i, k);
}
}
cout << ans << "\n";
return 0;
}
/*
1. Check borderline constraints. Can a variable you are dividing by be 0?
2. Use ll while using bitshifts
3. Do not erase from set while iterating it
4. Initialise everything
5. Read the task carefully, is something unique, sorted, adjacent, guaranteed??
6. DO NOT use if(!mp[x]) if you want to iterate the map later
7. Are you using i in all loops? Are the i's conflicting?
*/
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, A of length N, find the contiguous subarray within A which has the largest sum.First line of each test case contain the number of test cases.
The first line of each test case contains an integer n, the length of the array A
and the next line contains n integers.
Constraints:
1<=T<=100
1 <= N <= 10^5
-10^6 <= A[i] <= 10^6Output an integer representing the maximum possible sum of the contiguous subarray.Input:
1
5
1 2 3 4 -10
Output:
10
Explanation:-
1+2+3+4=10, I have written this Solution Code: t=int(input())
while t>0:
n=int(input())
a=map(int,input().split())
m=0
c=0
for i in a:
c+=i
if c>m:m=c
elif c<0:c=0
print(m)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, A of length N, find the contiguous subarray within A which has the largest sum.First line of each test case contain the number of test cases.
The first line of each test case contains an integer n, the length of the array A
and the next line contains n integers.
Constraints:
1<=T<=100
1 <= N <= 10^5
-10^6 <= A[i] <= 10^6Output an integer representing the maximum possible sum of the contiguous subarray.Input:
1
5
1 2 3 4 -10
Output:
10
Explanation:-
1+2+3+4=10, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
static long Sum(int a[], int size)
{
long max_so_far = -1000000007, max_ending_here = 0;
for (int i = 0; i < size; i++)
{
max_ending_here = max_ending_here + a[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
public static void main (String[] args) throws java.lang.Exception
{
BufferedReader br =
new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine().trim());
while(t-->0){
int n = Integer.parseInt(br.readLine().trim());
int arr[] = new int[n];
String inputLine[] = br.readLine().trim().split(" ");
for (int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(inputLine[i]);
}
System.out.println(Sum(arr,n));
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, A of length N, find the contiguous subarray within A which has the largest sum.First line of each test case contain the number of test cases.
The first line of each test case contains an integer n, the length of the array A
and the next line contains n integers.
Constraints:
1<=T<=100
1 <= N <= 10^5
-10^6 <= A[i] <= 10^6Output an integer representing the maximum possible sum of the contiguous subarray.Input:
1
5
1 2 3 4 -10
Output:
10
Explanation:-
1+2+3+4=10, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
long long Sum(long long a[], int size)
{
long long max_so_far = INT_MIN, max_ending_here = 0;
for (int i = 0; i < size; i++)
{
max_ending_here = max_ending_here + a[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
long long a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
cout<<Sum(a,n)<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: def findMin(arr, low, high):
if high < low:
return arr[0]
if high == low:
return arr[low]
mid = int((low + high)/2)
if mid < high and arr[mid+1] < arr[mid]:
return arr[mid+1]
if mid > low and arr[mid] < arr[mid - 1]:
return arr[mid]
if arr[high] > arr[mid]:
return findMin(arr, low, mid-1)
return findMin(arr, mid+1, high)
T =int(input())
for i in range(T):
N = int(input())
arr = list(input().split())
for k in range(len(arr)):
arr[k] = int(arr[k])
print(findMin(arr,0,N-1)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
int l = 0, h = n+1;
if(a[1] < a[n]){
cout << a[1] << endl;
continue;
}
while(l+1 < h){
int m = (l + h) >> 1;
if(a[m] >= a[1])
l = m;
else
h = m;
}
cout << a[h] << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main{
public static void main (String[] args) {
//code
Scanner s = new Scanner(System.in);
int t = s.nextInt();
for(int j=0;j<t;j++){
int al = s.nextInt();
int a[] = new int[al];
for(int i=0;i<al;i++){
a[i] = s.nextInt();
}
binSearchSmallest(a);
}
}
public static void binSearchSmallest(int a[]) {
int s=0;
int e = a.length - 1;
int mid = 0;
while(s<=e){
mid = (s+e)/2;
if(a[s]<a[e]){
System.out.println(a[s]);
return;
}
if(a[mid]>=a[s]){
s=mid+1;
}
else{
e=mid;
}
if(s == e){
System.out.println(a[s]);
return;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: // arr is the input array
function findMin(arr,n) {
// write code here
// do not console.log
// return the number
let min = arr[0]
for(let i=1;i<arr.length;i++){
if(min > arr[i]){
min = arr[i]
}
}
return min
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: John has N candies. He wants to crush all of them. He feels that it would be boring to crush the candies randomly, so he found a method to crush them. He divides these candies into a minimum number of groups such that no group contains more than 3 candies. He crushes one candy from each group. If there are G groups in a single step, then the cost incurred in crushing a single candy for that step is G dollars. After candy from each group is crushed, he takes all the remaining candies and repeats the process until he has no candies left. He hasn't started crushing yet, but he wants to know how much the total cost would be incurred. Can you help him?
You have to answer Q-independent queries.The first line of input contains a single integer, Q denoting the number of queries.
Next, Q lines contain a single integer N denoting the number of candies John has.
<b>Constraints</b>
1 <= Q <= 5 * 10^4
1 <= N <= 10^9Print Q lines containing total cost incurred for each query.Sample Input 1:
1
4
Sample Output 1:
6
<b>Explanation:</b>
Query 1: First step John divides the candies into two groups of 3 and 1 candy respectively. Crushing one-one candy from both groups would cost him 2x2 = 4 dollars. He is now left with 2 candies. He divides it into one group. He crushes one candy for 1 dollar. Now, he is left with 1 candy. He crushes the last candy for 1 dollar. So, the total cost incurred is 4+1+1 = 6 dollars., I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int cost(int n){
if(n == 0) return 0;
int g = (n-1)/3 + 1;
return g*g + cost(n-g);
}
signed main() {
IOS;
clock_t start = clock();
int q; cin >> q;
while(q--){
int n;
cin >> n;
cout << cost(n) << endl;
}
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: John has N candies. He wants to crush all of them. He feels that it would be boring to crush the candies randomly, so he found a method to crush them. He divides these candies into a minimum number of groups such that no group contains more than 3 candies. He crushes one candy from each group. If there are G groups in a single step, then the cost incurred in crushing a single candy for that step is G dollars. After candy from each group is crushed, he takes all the remaining candies and repeats the process until he has no candies left. He hasn't started crushing yet, but he wants to know how much the total cost would be incurred. Can you help him?
You have to answer Q-independent queries.The first line of input contains a single integer, Q denoting the number of queries.
Next, Q lines contain a single integer N denoting the number of candies John has.
<b>Constraints</b>
1 <= Q <= 5 * 10^4
1 <= N <= 10^9Print Q lines containing total cost incurred for each query.Sample Input 1:
1
4
Sample Output 1:
6
<b>Explanation:</b>
Query 1: First step John divides the candies into two groups of 3 and 1 candy respectively. Crushing one-one candy from both groups would cost him 2x2 = 4 dollars. He is now left with 2 candies. He divides it into one group. He crushes one candy for 1 dollar. Now, he is left with 1 candy. He crushes the last candy for 1 dollar. So, the total cost incurred is 4+1+1 = 6 dollars., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(
new InputStreamReader(System.in)
);
long q = Long.parseLong(br.readLine());
while(q-->0)
{
long N = Long.parseLong(br.readLine());
System.out.println(candyCrush(N,0,0));
}
}
static long candyCrush(long N, long cost,long group)
{
if(N==0)
{
return cost;
}
if(N%3==0)
{
group = N/3;
cost = cost + (group*group);
return candyCrush(N-group,cost,0);
}
else
{
group = (N/3)+1;
cost = cost + (group*group);
return candyCrush(N-group,cost,0);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <i>Your Grace, I feel I've been remiss in my duties. I've given you meat and wine and music, but I haven’t shown you the hospitality you deserve. My King has married and I owe my new Queen a wedding gift. </i>
Roose Bolton thinks of attacking Robb Stark, but since Catelyn has realised his traumatising idea, she signals Robb by shouting two numbers A and B. For Robb to understand the signal, he needs to solve the following problem with the two numbers:
Given A and B, find whether A can be transformed to B if A can be converted to A + fact(A) any number of times, where fact(A) is any factor of A other than 1 and A itself.
For example: A = 9, B = 14 can follow the given steps during conversion
A = 9
A = 9 + 3 = 12 (3 is a factor of 9)
A = 12 + 2 = 14 = B (2 is a factor of 12)The first and the only line of input contains two numbers A and B.
Constraints
1 <= A, B <= 10<sup>12</sup>Output "Yes" (without quotes) if we can transform A to B, else output "No" (without quotes)Sample Input
9 14
Sample Output
Yes
Explanation: In the problem statement
Sample Input
33 35
Sample Output
No
Explanation
The minimum factor of 33 is 3, so we cannot transform 33 to 35., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import java.util.StringTokenizer;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
};
public static void main(String[] args) {
FastReader sc = new FastReader();
PrintWriter pw = new PrintWriter(System.out);
long a=sc.nextLong();
long b=sc.nextLong();
long tmp=a;
res=new ArrayList<>();
factorize(a);
if(res.size()==1&&res.get(0)==a)
{
System.out.println("No");
return;
}
if(a==b)
{
System.out.println("Yes");
return;
}
for(long i=2;i <= (long) Math.sqrt(b);i++)
{
if(b%i==0&&(b-i)>=a)
{
for(long j:res)
{
if((b-i)%j==0)
{
System.out.println("Yes");
return;
}
}
}
}
System.out.println("No");
}
static ArrayList<Long> res;
static void factorize(long n) {
int count = 0;
while (!(n % 2 > 0)) {
n >>= 1;
count++;
}
if (count > 0) {
res.add(2l);
}
for (long i = 3; i <= (long) Math.sqrt(n); i += 2) {
count = 0;
while (n % i == 0) {
count++;
n = n / i;
}
if (count > 0) {
res.add(i);
}
}
if (n > 2) {
res.add(n);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <i>Your Grace, I feel I've been remiss in my duties. I've given you meat and wine and music, but I haven’t shown you the hospitality you deserve. My King has married and I owe my new Queen a wedding gift. </i>
Roose Bolton thinks of attacking Robb Stark, but since Catelyn has realised his traumatising idea, she signals Robb by shouting two numbers A and B. For Robb to understand the signal, he needs to solve the following problem with the two numbers:
Given A and B, find whether A can be transformed to B if A can be converted to A + fact(A) any number of times, where fact(A) is any factor of A other than 1 and A itself.
For example: A = 9, B = 14 can follow the given steps during conversion
A = 9
A = 9 + 3 = 12 (3 is a factor of 9)
A = 12 + 2 = 14 = B (2 is a factor of 12)The first and the only line of input contains two numbers A and B.
Constraints
1 <= A, B <= 10<sup>12</sup>Output "Yes" (without quotes) if we can transform A to B, else output "No" (without quotes)Sample Input
9 14
Sample Output
Yes
Explanation: In the problem statement
Sample Input
33 35
Sample Output
No
Explanation
The minimum factor of 33 is 3, so we cannot transform 33 to 35., I have written this Solution Code: import math
arr=[int (x) for x in input().split()]
a,b=arr[0],arr[1]
if a>=b:
print("No")
else:
m=-1
i=2
while i**2<=a:
if a%i==0:
m=i
break
i+=1
if m==-1:
print("No")
elif math.gcd(a,b)!=1:
print("Yes")
else:
a=a+m
if a<b and math.gcd(a,b)!=1:
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <i>Your Grace, I feel I've been remiss in my duties. I've given you meat and wine and music, but I haven’t shown you the hospitality you deserve. My King has married and I owe my new Queen a wedding gift. </i>
Roose Bolton thinks of attacking Robb Stark, but since Catelyn has realised his traumatising idea, she signals Robb by shouting two numbers A and B. For Robb to understand the signal, he needs to solve the following problem with the two numbers:
Given A and B, find whether A can be transformed to B if A can be converted to A + fact(A) any number of times, where fact(A) is any factor of A other than 1 and A itself.
For example: A = 9, B = 14 can follow the given steps during conversion
A = 9
A = 9 + 3 = 12 (3 is a factor of 9)
A = 12 + 2 = 14 = B (2 is a factor of 12)The first and the only line of input contains two numbers A and B.
Constraints
1 <= A, B <= 10<sup>12</sup>Output "Yes" (without quotes) if we can transform A to B, else output "No" (without quotes)Sample Input
9 14
Sample Output
Yes
Explanation: In the problem statement
Sample Input
33 35
Sample Output
No
Explanation
The minimum factor of 33 is 3, so we cannot transform 33 to 35., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
int small_prime(int x) {
for(int i=2; i*i<=x; i++){
if(x%i==0){
return i;
}
}
return -1;
}
void solve(){
int a, b;
cin>>a>>b;
int aa = a, bb = b;
int d1 = small_prime(a);
int d2 = small_prime(b);
if (d1 == -1 || d2 == -1) {
cout<<"No";
return;
}
if ((a % 2) == 1) {
a += d1;
}
if ((b % 2) == 1) {
b -= d2;
}
if (a > b) {
cout<<"No";
return;
}
cout<<"Yes";
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Mohit has three integers A, B, and C with him he wants to find the average of these three integers however he is weak in maths, so help him to find the average. You need to report the floor of the average value.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Average()</b> that takes integers A, B, and C as arguments.
Constraints:-
1 <= A, B, and C <= 10000Return the floor of average of these numbers.Sample Input:-
3 4 5
Sample Output:-
4
Sample Input:-
3 4 4
Sample Output:-
3, I have written this Solution Code: def Average(A,B,C):
return (A+B+C)//3
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Mohit has three integers A, B, and C with him he wants to find the average of these three integers however he is weak in maths, so help him to find the average. You need to report the floor of the average value.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Average()</b> that takes integers A, B, and C as arguments.
Constraints:-
1 <= A, B, and C <= 10000Return the floor of average of these numbers.Sample Input:-
3 4 5
Sample Output:-
4
Sample Input:-
3 4 4
Sample Output:-
3, I have written this Solution Code:
int Average(int A,int B, int C){
return (A+B+C)/3;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Mohit has three integers A, B, and C with him he wants to find the average of these three integers however he is weak in maths, so help him to find the average. You need to report the floor of the average value.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Average()</b> that takes integers A, B, and C as arguments.
Constraints:-
1 <= A, B, and C <= 10000Return the floor of average of these numbers.Sample Input:-
3 4 5
Sample Output:-
4
Sample Input:-
3 4 4
Sample Output:-
3, I have written this Solution Code:
static int Average(int A,int B, int C){
return (A+B+C)/3;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Mohit has three integers A, B, and C with him he wants to find the average of these three integers however he is weak in maths, so help him to find the average. You need to report the floor of the average value.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Average()</b> that takes integers A, B, and C as arguments.
Constraints:-
1 <= A, B, and C <= 10000Return the floor of average of these numbers.Sample Input:-
3 4 5
Sample Output:-
4
Sample Input:-
3 4 4
Sample Output:-
3, I have written this Solution Code:
int Average(int A,int B, int C){
return (A+B+C)/3;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of N integers. Compute the product of all of its elements. If the product exceeds 10<sup>18</sup>, return -1 instead.Since this will be a functional problem, you don't have to take input. You have to complete the function <b>findProduct()</b> that takes an ArrayList A as a parameter.
<b>Constraints</b>
1<=N<=10<sup>5</sup>
0<=A<sub>i</sub><=10<sup>18</sup>Return the product if it is less than or equal to 10<sup>18</sup. In all other cases, return -1.Sample Input:
10 5 3 2
Sample Output:
300, I have written this Solution Code:
class Solution {
public long findProduct(ArrayList<Long> A) {
long ans = 1;
for (long a : A) {
if (a == 0) {
ans = 0;
break;
} else if (a > (long)Math.pow(10, 18)/ans) {
ans = -1;
} else {
ans *= a;
}
}
return ans;
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: function Charity(n,m) {
// write code here
// do no console.log the answer
// return the output using return keyword
const per = Math.floor(m / n)
return per > 1 ? per : -1
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: static int Charity(int n, int m){
int x= m/n;
if(x<=1){return -1;}
return x;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: int Charity(int n, int m){
int x= m/n;
if(x<=1){return -1;}
return x;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: def Charity(N,M):
x = M//N
if x<=1:
return -1
return x
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: int Charity(int n, int m){
int x= m/n;
if(x<=1){return -1;}
return x;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Cirno's perfect bitmasks classroom has just started! Cirno gave her students a positive integer x. As an assignment, her students need to find the minimum positive integer y, which satisfies the following two conditions:
1. x and y>0
2. x xor y>0
Where and is the bitwise AND operation, and xor is the bitwise XOR operation.The input consists of an integer x.
<b>Constraints</b>
1 ≤ x ≤ 2<sup>30</sup>Print a single integer — the minimum number of y.<b>Sample Input 1</b>
1
<b>Sample Output 1</b>
3
<b>Sample Input 1</b>
5
<b>Sample Output 1</b>
1, I have written this Solution Code: #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
#define int long long int
#define endl '\n'
void fastio()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
using namespace __gnu_pbds;
const int M = 1e9 + 7;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
// Author:Abhas
void solution()
{
int n;
cin >> n;
int q = 0;
int k = __builtin_popcount(n);
for (int i = 0; i < 31; i++)
{
if ((1 << i) & n)
{
q = 1 << i;
break;
}
}
if (k == 1)
{
if (q % 2)
{
q += 2;
}
else
{
q++;
}
cout << q << endl;
}
else
{
cout << q << endl;
}
}
signed main()
{
fastio();
int t = 1;
// cin >> t;
while (t--)
{
solution();
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: def DivisorProblem(N):
ans=0
while N>1:
cnt=2
while N%cnt!=0:
cnt=cnt+1
N = N//cnt
ans=ans+1
return ans
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: static int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a sequence of length N consisting of integers:
A=(A<sub>1</sub>, A<sub>2</sub>,...., A<sub>N</sub>)
Find the smallest non-negative integer not in (A<sub>1</sub>, A<sub>2</sub>,....,A<sub>N</sub>).The input contains N and elements of sequence separated by a new line.
N
A<sub>1</sub>, A<sub>2</sub>,. , A<sub>N</sub>
<b>Constraints</b>
1≤N≤2000
0≤Ai≤2000
All values in the input are integers.Print the answer.<b>Sample Input 1</b>
8
0 3 2 6 2 1 0 0
<b>Sample Output 1</b>
4
<b>Sample Input 2</b>
3
2000 2000 2000
<b>Sample Output 2</b>
0, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
using LL = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int N;
cin >> N;
set<int> s;
for (int i = 0, A; i < N; i += 1) {
cin >> A;
s.insert(A);
}
int ans = 0;
while (s.count(ans)) ans += 1;
cout << ans;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an 8*8 empty chessboard in which a knight is placed at a position (X, Y). Your task is to find the number of positions in the chessboard knight can jump into in a single move .
Note:- Rows and Columns are numbered through 1 to N.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Knight()</b> that takes integers X and Y as arguments.
Constraints:-
1 <= X <= 8
1 <= Y <= 8Return the number of positions Knight can jump into in a single move.Sample input:-
4 5
Sample Output:-
8
Explanation:-
Positions:- (3, 3), (5, 3), (3, 7), (5, 7), (6, 6), (6, 4), (2, 6), (2, 4)
Sample input:-
1 1
Sample Output:-
2
Explanation:-
Positions:- (3, 2), (2, 3), I have written this Solution Code: def Knight(X,Y):
cnt=0
if(X>2):
if(Y>1):
cnt=cnt+1
if(Y<8):
cnt=cnt+1
if(Y>2):
if(X>1):
cnt=cnt+1
if(X<8):
cnt=cnt+1
if(X<7):
if(Y>1):
cnt=cnt+1
if(Y<8):
cnt=cnt+1
if(Y<7):
if(X>1):
cnt=cnt+1
if(X<8):
cnt=cnt+1
return cnt;
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an 8*8 empty chessboard in which a knight is placed at a position (X, Y). Your task is to find the number of positions in the chessboard knight can jump into in a single move .
Note:- Rows and Columns are numbered through 1 to N.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Knight()</b> that takes integers X and Y as arguments.
Constraints:-
1 <= X <= 8
1 <= Y <= 8Return the number of positions Knight can jump into in a single move.Sample input:-
4 5
Sample Output:-
8
Explanation:-
Positions:- (3, 3), (5, 3), (3, 7), (5, 7), (6, 6), (6, 4), (2, 6), (2, 4)
Sample input:-
1 1
Sample Output:-
2
Explanation:-
Positions:- (3, 2), (2, 3), I have written this Solution Code: static int Knight(int X, int Y){
int cnt=0;
if(X>2){
if(Y>1){cnt++;}
if(Y<8){cnt++;}
}
if(Y<7){
if(X>1){cnt++;}
if(X<8){cnt++;}
}
if(X<7){
if(Y>1){cnt++;}
if(Y<8){cnt++;}
}
if(Y>2){
if(X>1){cnt++;}
if(X<8){cnt++;}
}
return cnt;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an 8*8 empty chessboard in which a knight is placed at a position (X, Y). Your task is to find the number of positions in the chessboard knight can jump into in a single move .
Note:- Rows and Columns are numbered through 1 to N.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Knight()</b> that takes integers X and Y as arguments.
Constraints:-
1 <= X <= 8
1 <= Y <= 8Return the number of positions Knight can jump into in a single move.Sample input:-
4 5
Sample Output:-
8
Explanation:-
Positions:- (3, 3), (5, 3), (3, 7), (5, 7), (6, 6), (6, 4), (2, 6), (2, 4)
Sample input:-
1 1
Sample Output:-
2
Explanation:-
Positions:- (3, 2), (2, 3), I have written this Solution Code: int Knight(int X, int Y){
int cnt=0;
if(X>2){
if(Y>1){cnt++;}
if(Y<8){cnt++;}
}
if(Y<7){
if(X>1){cnt++;}
if(X<8){cnt++;}
}
if(X<7){
if(Y>1){cnt++;}
if(Y<8){cnt++;}
}
if(Y>2){
if(X>1){cnt++;}
if(X<8){cnt++;}
}
return cnt;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an 8*8 empty chessboard in which a knight is placed at a position (X, Y). Your task is to find the number of positions in the chessboard knight can jump into in a single move .
Note:- Rows and Columns are numbered through 1 to N.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Knight()</b> that takes integers X and Y as arguments.
Constraints:-
1 <= X <= 8
1 <= Y <= 8Return the number of positions Knight can jump into in a single move.Sample input:-
4 5
Sample Output:-
8
Explanation:-
Positions:- (3, 3), (5, 3), (3, 7), (5, 7), (6, 6), (6, 4), (2, 6), (2, 4)
Sample input:-
1 1
Sample Output:-
2
Explanation:-
Positions:- (3, 2), (2, 3), I have written this Solution Code: int Knight(int X, int Y){
int cnt=0;
if(X>2){
if(Y>1){cnt++;}
if(Y<8){cnt++;}
}
if(Y<7){
if(X>1){cnt++;}
if(X<8){cnt++;}
}
if(X<7){
if(Y>1){cnt++;}
if(Y<8){cnt++;}
}
if(Y>2){
if(X>1){cnt++;}
if(X<8){cnt++;}
}
return cnt;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of integers, find the number of subarrays with an odd sum.First line contains an integers N.
Next line contains N space separated integers denoting elements of array.
Constraints
1 <= N <= 10^5
1 <= Ai <= 10^5Print the number of subarrays with an odd sum.Sample Input 1:
3
1 3 5
Output
4
Explanation:
All subarrays are [1], [1, 3], [1, 3, 5], [3], [3, 5], [5]
All sub- arrays sum are [1, 4, 9, 3, 8, 5].
Odd sums are [1, 9, 3, 5] so the answer is 4.
Sample Input 2:
3
2 4 6
Output
0
Explanation:
All subarrays are [2], [2, 4], [2, 4, 6], [4], [4, 6], [6]
All sub- arrays sum are [2, 6, 12, 4, 10, 6].
All sub- arrays have even sum and the answer is 0., I have written this Solution Code:
def countOddSum(a, n):
# 'odd' stores number of
# odd numbers upto ith index
# 'c_odd' stores number of
# odd sum subarrays starting
# at ith index
# 'Result' stores the number
# of odd sum subarrays
c_odd = 0;
result = 0;
odd = False;
# First find number of odd
# sum subarrays starting at
# 0th index
for i in range(n):
if (a[i] % 2 == 1):
if(odd == True):
odd = False;
else:
odd = True;
if (odd):
c_odd += 1;
# Find number of odd sum
# subarrays starting at ith
# index add to result
for i in range(n):
result += c_odd;
if (a[i] % 2 == 1):
c_odd = (n - i - c_odd);
return result;
n=int(input())
arr=list(map(int,input().strip().split()))[:n]
print(countOddSum(arr,n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of integers, find the number of subarrays with an odd sum.First line contains an integers N.
Next line contains N space separated integers denoting elements of array.
Constraints
1 <= N <= 10^5
1 <= Ai <= 10^5Print the number of subarrays with an odd sum.Sample Input 1:
3
1 3 5
Output
4
Explanation:
All subarrays are [1], [1, 3], [1, 3, 5], [3], [3, 5], [5]
All sub- arrays sum are [1, 4, 9, 3, 8, 5].
Odd sums are [1, 9, 3, 5] so the answer is 4.
Sample Input 2:
3
2 4 6
Output
0
Explanation:
All subarrays are [2], [2, 4], [2, 4, 6], [4], [4, 6], [6]
All sub- arrays sum are [2, 6, 12, 4, 10, 6].
All sub- arrays have even sum and the answer is 0., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main
{
public static void main (String args[]) throws IOException
{
BufferedReader br = new BufferedReader (new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
long a[] = new long[n];
String line = br.readLine(); // to read multiple integers line
String[] strs = line.trim().split("\\s+");
for (int i = 0; i < n; i++) {
a[i] = Long.parseLong(strs[i]);
}
long[] prefix=new long[n];
prefix[0]=a[0];
for(int i=1;i<n;i++){
prefix[i]=prefix[i-1]+a[i];
}
long e=1;
long o=0;
for(int i=0;i<n;i++){
if(prefix[i]%2==0)e++;
else o++;
}
e*=o;
System.out.print(e);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integer N and K, find the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).
Where X = Summation (2<sup>(4*i)</sup>) for 1 <= i <= 25.The first and the only line of input contains two space separated integers N and K.
Constraints
1 <= N <= 10^18
1 <= K <= 10^18Print a single integer which is the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).Sample Input 1
40 4
Sample Output 1
1
Sample Input
40 3
Sample Output
0, I have written this Solution Code: import java.util.InputMismatchException;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
public class Main {
InputStream is;
PrintWriter out;
String INPUT = "";
int MAX = (int) 1e5, MOD = (int)1e9+7;
void solve(int TC) {
long n = nl();
long k = nl();
int p = 0;
while(n>0 && n%2==0) {
n/=2;
++p;
}
if(p>=k) {pn(0);return;}
k -= p;
long ans = (k+3L)/4L;
pn(ans);
}
boolean TestCases = false;
public static void main(String[] args) throws Exception { new Main().run(); }
long pow(long a, long b) {
if(b==0 || a==1) return 1;
long o = 1;
for(long p = b; p > 0; p>>=1) {
if((p&1)==1) o = (o*a) % MOD;
a = (a*a) % MOD;
} return o;
}
long inv(long x) {
long o = 1;
for(long p = MOD-2; p > 0; p>>=1) {
if((p&1)==1)o = (o*x)%MOD;
x = (x*x)%MOD;
} return o;
}
long gcd(long a, long b) { return (b==0) ? a : gcd(b,a%b); }
int gcd(int a, int b) { return (b==0) ? a : gcd(b,a%b); }
void run() throws Exception {
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
int T = TestCases ? ni() : 1;
for(int t=1;t<=T;t++) solve(t);
out.flush();
if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
}
void p(Object o) { out.print(o); }
void pn(Object o) { out.println(o); }
void pni(Object o) { out.println(o);out.flush(); }
double PI = 3.141592653589793238462643383279502884197169399;
int ni() {
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-') {
minus = true;
b = readByte();
}
while(true) {
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
long nl() {
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-') {
minus = true;
b = readByte();
}
while(true) {
if(b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
double nd() { return Double.parseDouble(ns()); }
char nc() { return (char)skip(); }
int BUF_SIZE = 1024 * 8;
byte[] inbuf = new byte[BUF_SIZE];
int lenbuf = 0, ptrbuf = 0;
int readByte() {
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
} return inbuf[ptrbuf++];
}
boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
String ns() {
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))) {
sb.appendCodePoint(b); b = readByte();
} return sb.toString();
}
char[] ns(int n) {
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
b = readByte();
} return n == p ? buf : Arrays.copyOf(buf, p);
}
void tr(Object... o) { if(INPUT.length() > 0)System.out.println(Arrays.deepToString(o)); }
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integer N and K, find the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).
Where X = Summation (2<sup>(4*i)</sup>) for 1 <= i <= 25.The first and the only line of input contains two space separated integers N and K.
Constraints
1 <= N <= 10^18
1 <= K <= 10^18Print a single integer which is the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).Sample Input 1
40 4
Sample Output 1
1
Sample Input
40 3
Sample Output
0, I have written this Solution Code: [n,k]=[int(j) for j in input().split()]
a=0
while n%2==0:
a+=1
n=n//2
if k>a:
print((k-a-1)//4+1)
else:
print(0), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integer N and K, find the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).
Where X = Summation (2<sup>(4*i)</sup>) for 1 <= i <= 25.The first and the only line of input contains two space separated integers N and K.
Constraints
1 <= N <= 10^18
1 <= K <= 10^18Print a single integer which is the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).Sample Input 1
40 4
Sample Output 1
1
Sample Input
40 3
Sample Output
0, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n,k;
cin>>n>>k;
while(k&&n%2==0){
n/=2;
--k;
}
cout<<(k+3)/4;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b, your task is to check following conditions:-
1. If a <= 10 and b >= 10 (Logical AND).
2. Atleast one from a or b will be even (Logical OR).
3. if a is not equal to b (Logical NOT).The first line of the input contains 2 integers a and b.
<b>Constraints:</b>
1 <= a, b <= 100Print the string <b>"true"</b> if the condition holds in each function else <b>"false"</b> .
Sample Input:-
3 12
Sample Output:-
true true true
Explanation
So a = 3 and b = 12, so a<=10 and b>=10 hence first condition true, a is not even but b is even so atleast one of them is even hence true, third a != b which is also true hence the final output comes true true true.
Sample Input:-
10 10
Sample Output:-
true true false
, I have written this Solution Code: a, b = list(map(int, input().split(" ")))
print(str(a <= 10 and b >= 10).lower(), end=' ')
print(str(a % 2 == 0 or b % 2 == 0).lower(), end=' ')
print(str(not a == b).lower()), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b, your task is to check following conditions:-
1. If a <= 10 and b >= 10 (Logical AND).
2. Atleast one from a or b will be even (Logical OR).
3. if a is not equal to b (Logical NOT).The first line of the input contains 2 integers a and b.
<b>Constraints:</b>
1 <= a, b <= 100Print the string <b>"true"</b> if the condition holds in each function else <b>"false"</b> .
Sample Input:-
3 12
Sample Output:-
true true true
Explanation
So a = 3 and b = 12, so a<=10 and b>=10 hence first condition true, a is not even but b is even so atleast one of them is even hence true, third a != b which is also true hence the final output comes true true true.
Sample Input:-
10 10
Sample Output:-
true true false
, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
class Main {
static boolean Logical_AND(int a, int b){
if(a<=10 && b>=10){
return true;}
return false;}
static boolean Logical_OR(int a, int b){
if(a%2==0 || b%2==0){
return true;}
return false;}
static boolean Logical_NOT(int a, int b){
if(a!=b){
return true;}
return false;}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int a=in.nextInt();
int b=in.nextInt();
System.out.print(Logical_AND(a, b)+" ");
System.out.print(Logical_OR(a,b)+" ");
System.out.print(Logical_NOT(a,b)+" ");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given 2 parameters: a low and high number. Your goal is to print all numbers between low and high,
and for each of these numbers print whether or not the number is divisible by 3. If the number is divisible by 3,
print the word "div3" directly after the number.2 numbers, one will be low and other high.
0<=low<=high<=10000If the number is divisible by 3, print the word "div3" directly after the number.Sample input:-
1 6
Sample output:-
1 2 3 div3 4 5 6 div3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int low = sc.nextInt();
int high = sc.nextInt();
for(int i = low; i <= high; i++){
if(i%3 == 0){
System.out.print(i);
System.out.print(" ");
System.out.print("div"+3);
System.out.print(" ");
}
else{
System.out.print(i);
System.out.print(" ");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given 2 parameters: a low and high number. Your goal is to print all numbers between low and high,
and for each of these numbers print whether or not the number is divisible by 3. If the number is divisible by 3,
print the word "div3" directly after the number.2 numbers, one will be low and other high.
0<=low<=high<=10000If the number is divisible by 3, print the word "div3" directly after the number.Sample input:-
1 6
Sample output:-
1 2 3 div3 4 5 6 div3, I have written this Solution Code: inp = input("").split(" ")
init = []
for i in range(int(inp[0]),int(inp[1])+1):
if(i%3 == 0):
init.append(str(i)+" div3")
else:
init.append(str(i))
print(" ".join(init)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given 2 parameters: a low and high number. Your goal is to print all numbers between low and high,
and for each of these numbers print whether or not the number is divisible by 3. If the number is divisible by 3,
print the word "div3" directly after the number.2 numbers, one will be low and other high.
0<=low<=high<=10000If the number is divisible by 3, print the word "div3" directly after the number.Sample input:-
1 6
Sample output:-
1 2 3 div3 4 5 6 div3, I have written this Solution Code: function test_divisors(low, high) {
// we'll store all numbers and strings within an array
// instead of printing directly to the console
const output = [];
for (let i = low; i <= high; i++) {
// simply store the current number in the output array
output.push(i);
// check if the current number is evenly divisible by 3
if (i % 3 === 0) { output.push('div3'); }
}
// return all numbers and strings
console.log(output.join(" "));
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: John is confused about the number of ways to propose to Olivia. So, he asked for your help. Now, you need to determine the number of ways for John to propose to Olivia.
For that, You are given an integer N and you need to count the number of pairs (x<sub>1</sub>, x<sub>2</sub>) such that x<sub>1</sub><sup>2</sup> + x<sub>2</sub><sup>2</sup> = N and x<sub>1</sub>, x<sub>2</sub> both are positive integer.The first line contains a single integer T, the number of test cases.
T lines follow. Each line describes a single test case and contains a single integer N.
<b>Constraints:</b>
1 <= T <= 100
2 <= N <= 10<sup>5</sup>For each test case, print a single integer count of such pairs.Sample Input 1:
2
13
4
Sample Output 2:
2
0, I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC target("popcnt")
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
using cd = complex<double>;
const double PI = acos(-1);
// DEFINE STATEMENTS
const long long infty = 1e18;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define pob pop_back
#define fr first
#define sc second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
typedef long long ll;
typedef long long unsigned int llu;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
// DEBUG FUNCTIONS
#ifndef ONLINE_JUDGE
template<typename T>
void __p(T a) {
cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
cout<<"{";
__p(a.first);
cout<<",";
__p(a.second);
cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
cout<<"{";
for(auto it=a.begin(); it<a.end(); it++)
__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T>
void __p(std::multiset<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
cout<<"{\n";
for(auto it=a.begin(); it!=a.end();++it)
{
__p(it->first);
cout << ": ";
__p(it->second);
cout<<"\n";
}
cout << "}\n";
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
__p(a1);
__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
cout<<name<<" : ";
__p(arg1);
cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
int bracket=0,i=0;
for(;; i++)
if(names[i]==','&&bracket==0)
break;
else if(names[i]=='(')
bracket++;
else if(names[i]==')')
bracket--;
const char *comma=names+i;
cout.write(names,comma-names)<<" : ";
__p(arg1);
cout<<" | ";
__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif
void solve(){
ll n; cin >> n;
ll ans = 0;
for(ll i = 1;i*i<n;i++){
ll x = sqrt(n - i*i);
if(x*x + i*i == n){
ans++;
}
}
cout << ans << "\n";
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
// cout.tie(NULL);
#ifdef LOCALFLAG
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
ll t = 1;
cin >> t;
while(t--){
solve();
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Create a class named 'Student' with String variable 'name' and integer variable 'rollNumber'.
You need to perform the below operations in <b>myFunction()</b>:
<ul>
<li>Assign the value of <b>rollNumber</b> as given by the user</li>
<li>Assign the value of <b>name</b> as given by the user</li>
</ul>The input contains a single line of String and integer value separated by space.You just have to assign values to Student class attributes. The driver code is handling the outputInput:
Gaurav 1
Output:
Gaurav 1
Input:
Swapnil 2
Output:
Swapnil 2, I have written this Solution Code:
class Student {
String name;
int rollNumber;
public void myFunction (String name, int rollNumber){
this.name = name;
this.rollNumber = rollNumber;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Create a class named 'Student' with String variable 'name' and integer variable 'rollNumber'.
You need to perform the below operations in <b>myFunction()</b>:
<ul>
<li>Assign the value of <b>rollNumber</b> as given by the user</li>
<li>Assign the value of <b>name</b> as given by the user</li>
</ul>The input contains a single line of String and integer value separated by space.You just have to assign values to Student class attributes. The driver code is handling the outputInput:
Gaurav 1
Output:
Gaurav 1
Input:
Swapnil 2
Output:
Swapnil 2, I have written this Solution Code: class Student:
def __init__(self, name, roll_no):
self.name, self.roll_no = name ,roll_no
def myFunction(name, roll_no):
obj = Student(name, roll_no)
return obj, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
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